If $\alpha$ and $\beta$ are the zeroes of $f(x) = 2x^2 + 5x + k$ such that $\alpha^2 + \beta^2 + \alpha \beta = \dfrac{21}{4}$, find $k$.
Step-by-Step Solution
Key Concept: $\alpha + \beta = -5/2, \alpha \beta = k/2$. $(\alpha + \beta)^2 - \alpha \beta = 21/4$.
$\alpha + \beta = -5/2, \alpha \beta = k/2$. [0.5 Mark]
$(-5/2)^2 - k/2 = 21/4 \Rightarrow 25/4 - k/2 = 21/4$. [1.5 Marks]
$k/2 = 4/4 = 1 \Rightarrow k = 2$. [1.0 Mark]
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🎯 Official CBSE Marking Scheme:
Finding sum and product: 0.5 Mark
Substituting into expression $25/4 - k/2 = 21/4$: 1.5 Marks
Solving $k = 2$: 1.0 Mark
Correct Answer: